Matrices and Systems of Equations: Practice

The matrix is already triangular. The bottom row gives $3z = 9$, so $z = 3$. The middle row gives $y + z = 5$.

$y + 3 = 5 \;\Rightarrow\; y = 2$

The top row gives $x + y + z = 6$, so $x + 2 + 3 = 6$ and $x = 1$. The solution is $(1, 2, 3)$.

The zero row gives no constraint, so $z$ is free; call it $t$. The second row gives $y - z = 1$, so $y = 1 + t$. The first gives $x + 2z = 5$.

$x = 5 - 2t,\quad y = 1 + t,\quad z = t$

Every value of $t$ yields a solution, so there are infinitely many.

The bottom row reads $0x + 0y + 0z = 5$, i.e. $0 = 5$.

$\left[\begin{array}{ccc|c}0 & 0 & 0 & 5\end{array}\right] \Rightarrow 0 = 5$

That is impossible, so the system is inconsistent: no solution, regardless of the other rows.

The zero row adds nothing, and no contradiction appears, so the system is consistent. The variable $y$ has no pivot, so it is free.

$x = 3 - 2y,\quad z = 4,\quad y \text{ free}$

A free variable means infinitely many solutions.

The top-left entry is $0$, so swap the rows to get $\left[\begin{array}{cc|c}3 & 1 & 9 \\ 0 & 2 & 6\end{array}\right]$. The bottom row gives $2y = 6$, so $y = 3$. Then $3x + y = 9$.

$3x + 3 = 9 \;\Rightarrow\; x = 2$

So $(x, y) = (2, 3)$.

Apply $R_2 \to R_2 - 2R_1$: the second row becomes $(0,\ k - 6 \mid -6)$. A unique solution needs a pivot in the second column, so $k - 6 \ne 0$.

$k \ne 6$

When $k = 6$ the row reads $0 = -6$, giving no solution instead.

Each variable with a pivot is determined; the rest are free. With $3$ variables and $2$ pivots,

$3 - 2 = 1$

there is $1$ free variable, so the system has infinitely many solutions.

A pivot cannot be $0$. If a nonzero entry sits below, swapping brings it up so reduction can proceed.

$R_i \leftrightarrow R_j$

The swap is legal and changes nothing about the solution set.

The system is $x + y = 1$, $3x - y = -5$. Use $R_2 \to R_2 - 3R_1$ to get $(0,\ -4 \mid -8)$, so $-4y = -8$ and $y = 2$. Back-substitute into $x + y = 1$.

$x + 2 = 1 \;\Rightarrow\; x = -1$

So $(x, y) = (-1, 2)$.

Three pivots for three variables determine each one by back-substitution, and no contradiction row appears.

$3 \text{ pivots for } 3 \text{ variables} \Rightarrow \text{unique solution}$

The entry $m - 4$ is a pivot only when it is nonzero. It fails to give a pivot when

$m - 4 = 0 \;\Rightarrow\; m = 4$

Then the row is all zeros and $z$ is free.

Row $2$ has a leading $1$, so swapping it to the top avoids dividing row $1$ by $4$.

$R_1 \leftrightarrow R_2$

A leading $1$ keeps every later entry an integer.